Proof.
If (1), then (4) as a right adjoint for M \mapsto M \otimes _ A K is N \mapsto \mathop{\mathrm{Hom}}\nolimits _ B(K, N), see Differential Graded Algebra, Lemma 22.30.3. If (4), then (3) by Categories, Lemma 4.24.5. The implication (3) \Rightarrow (2) is immediate from the definitions.
Assume (2). We will prove (1). By the discussion in Homology, Section 12.7 the functor F is additive. Hence F induces a ring map A \to \text{End}_ B(F(M)), a \mapsto F(a \cdot \text{id}_ M) for every A-module M. We conclude that F(M) is an A \otimes _\mathbf {Z} B-module functorially in M. Set K = F(A). Define
M \otimes _ A K = M \otimes _ A F(A) \longrightarrow F(M), \quad m \otimes k \longmapsto F(\varphi _ m)(k)
Here \varphi _ m : A \to M sends a \to am. The rule (m, k) \mapsto F(\varphi _ m)(k) is A-bilinear (and B-linear on the right) as required to obtain the displayed A \otimes _\mathbf {Z} B-linear map. This construction is functorial in M, hence defines a transformation of functors - \otimes _ A K \to F(-) which is an isomorphism when evaluated on A. For every A-module M we can choose an exact sequence
\bigoplus \nolimits _{j \in J} A \to \bigoplus \nolimits _{i \in I} A \to M \to 0
Using the maps constructed above we find a commutative diagram
\xymatrix{ (\bigoplus \nolimits _{j \in J} A) \otimes _ A K \ar[r] \ar[d] & (\bigoplus \nolimits _{i \in I} A) \otimes _ A K \ar[r] \ar[d] & M \otimes _ A K \ar[r] \ar[d] & 0 \\ F(\bigoplus \nolimits _{j \in J} A) \ar[r] & F(\bigoplus \nolimits _{i \in I} A) \ar[r] & F(M) \ar[r] & 0 }
The lower row is exact as F is right exact. The upper row is exact as tensor product with K is right exact. Since F commutes with direct sums the left two vertical arrows are bijections. Hence we conclude.
\square
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